The Capacitance Formula gives the relationship between the charge stored on a conductor and the electric potential difference across it, expressed as \( C = Q/V \). This formula is a cornerstone of Class 12 Physics (NCERT Chapter 2 — Electrostatic Potential and Capacitance) and appears consistently in CBSE board exams, JEE Main, JEE Advanced, and NEET. In this article, you will find the complete derivation, a full formula sheet, three progressive solved examples, common mistakes, and expert exam tips for 2025-26.
Key Capacitance Formulas at a Glance
Quick reference for the most important capacitance formulas used in CBSE and competitive exams.
- Basic definition: \( C = \dfrac{Q}{V} \)
- Parallel plate capacitor: \( C = \dfrac{\varepsilon_0 A}{d} \)
- With dielectric: \( C = \dfrac{K \varepsilon_0 A}{d} \)
- Series combination: \( \dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} \)
- Parallel combination: \( C_{eq} = C_1 + C_2 + C_3 \)
- Energy stored: \( U = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C} \)
- Spherical capacitor: \( C = 4\pi\varepsilon_0 r \)
What is the Capacitance Formula?
The Capacitance Formula defines the ability of a conductor or a system of conductors to store electric charge per unit potential difference. A capacitor stores electrical energy in the electric field between its plates. The greater the capacitance, the more charge a capacitor can store at a given voltage.
In NCERT Class 12 Physics, Chapter 2 (Electrostatic Potential and Capacitance), capacitance is introduced as a fundamental electrical property. It is measured in Farads (F) in the SI system. One Farad equals one Coulomb per Volt.
Capacitance depends on the geometry of the conductors and the medium between them. It does not depend on the charge placed on the capacitor or the potential to which it is raised. This makes it a purely geometric and material property. Understanding the Capacitance Formula is essential for topics like energy storage, RC circuits, and dielectrics.
The official NCERT textbook for Class 12 Physics is available on the NCERT official website for free download.
Capacitance Formula — Expression and Variables
The fundamental Capacitance Formula is:
\[ C = \frac{Q}{V} \]
where C is capacitance, Q is the charge stored, and V is the potential difference across the capacitor.
| Symbol | Quantity | SI Unit |
|---|---|---|
| C | Capacitance | Farad (F) |
| Q | Electric charge stored | Coulomb (C) |
| V | Electric potential difference | Volt (V) |
| ϵ0 | Permittivity of free space | F m−1 (8.85 × 10−12 F/m) |
| A | Area of each plate (parallel plate capacitor) | m² |
| d | Distance between plates | metre (m) |
| K | Dielectric constant (relative permittivity) | Dimensionless |
| r | Radius of spherical conductor | metre (m) |
Derivation of the Parallel Plate Capacitor Formula
Consider two parallel conducting plates, each of area A, separated by a distance d in vacuum.
Step 1: The electric field between the plates is \( E = \dfrac{\sigma}{\varepsilon_0} \), where \( \sigma = Q/A \) is the surface charge density.
Step 2: The potential difference between the plates is \( V = E \cdot d = \dfrac{\sigma d}{\varepsilon_0} = \dfrac{Qd}{\varepsilon_0 A} \).
Step 3: Apply the definition \( C = Q/V \):
\[ C = \frac{Q}{V} = \frac{Q}{\dfrac{Qd}{\varepsilon_0 A}} = \frac{\varepsilon_0 A}{d} \]
When a dielectric of constant K fills the gap, the formula becomes \( C = K\varepsilon_0 A / d \). This shows that capacitance increases by a factor K when a dielectric is inserted.
Complete Capacitance Formula Sheet
The table below covers all major capacitance-related formulas for CBSE Class 12, JEE, and NEET preparation.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Basic Capacitance | \( C = Q/V \) | Q = charge, V = potential difference | Farad (F) | Class 12, Ch 2 |
| Parallel Plate (vacuum) | \( C = \varepsilon_0 A / d \) | A = plate area, d = separation | Farad (F) | Class 12, Ch 2 |
| Parallel Plate (with dielectric) | \( C = K\varepsilon_0 A / d \) | K = dielectric constant | Farad (F) | Class 12, Ch 2 |
| Spherical Capacitor (isolated sphere) | \( C = 4\pi\varepsilon_0 r \) | r = radius of sphere | Farad (F) | Class 12, Ch 2 |
| Cylindrical Capacitor | \( C = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)} \) | L = length, a = inner radius, b = outer radius | Farad (F) | Class 12, Ch 2 |
| Series Combination | \( \dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} \) | C1, C2, C3 = individual capacitances | Farad (F) | Class 12, Ch 2 |
| Parallel Combination | \( C_{eq} = C_1 + C_2 + C_3 \) | C1, C2, C3 = individual capacitances | Farad (F) | Class 12, Ch 2 |
| Energy Stored in Capacitor | \( U = \dfrac{1}{2}CV^2 \) | C = capacitance, V = voltage | Joule (J) | Class 12, Ch 2 |
| Energy in terms of Q | \( U = \dfrac{Q^2}{2C} \) | Q = charge, C = capacitance | Joule (J) | Class 12, Ch 2 |
| Energy Density | \( u = \dfrac{1}{2}\varepsilon_0 E^2 \) | E = electric field between plates | J/m³ | Class 12, Ch 2 |
| Charge on Capacitor | \( Q = CV \) | C = capacitance, V = voltage | Coulomb (C) | Class 12, Ch 2 |
| Effect of Dielectric on Energy | \( U’ = U/K \) (charge constant) | K = dielectric constant, U = initial energy | Joule (J) | Class 12, Ch 2 |
Capacitance Formula — Solved Examples
Example 1 (Class 10-11 Level): Basic Capacitance Calculation
Problem: A capacitor stores a charge of 120 μC when connected to a 12 V battery. Find the capacitance of the capacitor.
Given: Q = 120 μC = 120 × 10−6 C, V = 12 V
Step 1: Write the Capacitance Formula: \( C = \dfrac{Q}{V} \)
Step 2: Substitute the given values:
\[ C = \frac{120 \times 10^{-6}}{12} = 10 \times 10^{-6} \text{ F} \]
Step 3: Convert to microfarads: C = 10 μF
Answer
The capacitance of the capacitor is 10 μF.
Example 2 (Class 12 Level): Parallel Plate Capacitor with Dielectric
Problem: A parallel plate capacitor has plate area 6 × 10−3 m² and plate separation 3 mm. A dielectric slab of dielectric constant K = 4 is inserted between the plates. Calculate the capacitance. Take \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m.
Given: A = 6 × 10−3 m², d = 3 mm = 3 × 10−3 m, K = 4, ϵ0 = 8.85 × 10−12 F/m
Step 1: Use the formula for a parallel plate capacitor with dielectric:
\[ C = \frac{K\varepsilon_0 A}{d} \]
Step 2: Substitute all values:
\[ C = \frac{4 \times 8.85 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} \]
Step 3: Calculate numerator: \( 4 \times 8.85 \times 10^{-12} \times 6 \times 10^{-3} = 212.4 \times 10^{-15} \)
Step 4: Divide by d: \( C = \dfrac{212.4 \times 10^{-15}}{3 \times 10^{-3}} = 70.8 \times 10^{-12} \) F
Answer
The capacitance with dielectric is 70.8 pF.
Example 3 (JEE/NEET Level): Energy Stored and Combination of Capacitors
Problem: Three capacitors of capacitances 2 μF, 3 μF, and 6 μF are connected in series across a 90 V supply. Find (a) the equivalent capacitance, (b) the charge on each capacitor, and (c) the total energy stored in the system.
Given: C1 = 2 μF, C2 = 3 μF, C3 = 6 μF, V = 90 V
Step 1: Apply the series combination formula:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \]
Step 2: Find the LCM and add: \( \dfrac{1}{C_{eq}} = \dfrac{3 + 2 + 1}{6} = \dfrac{6}{6} = 1 \), so \( C_{eq} = 1 \) μF
Step 3: In series, the same charge appears on each capacitor:
\[ Q = C_{eq} \times V = 1 \times 10^{-6} \times 90 = 90 \times 10^{-6} \text{ C} = 90 \text{ \mu C} \]
Step 4: Calculate total energy stored:
\[ U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times 1 \times 10^{-6} \times (90)^2 \]
\[ U = \frac{1}{2} \times 10^{-6} \times 8100 = 4050 \times 10^{-6} \text{ J} = 4.05 \text{ mJ} \]
Answer
(a) Equivalent capacitance = 1 μF
(b) Charge on each capacitor = 90 μC
(c) Total energy stored = 4.05 mJ
CBSE Exam Tips 2025-26
- Memorise the unit relationship: 1 Farad = 1 Coulomb/Volt. CBSE frequently asks for unit derivation in 1-mark questions in 2025-26 board exams.
- Learn both series and parallel rules: Series capacitance behaves opposite to series resistance. We recommend writing this comparison in your notes to avoid confusion.
- Dielectric problems are high-weightage: CBSE 2025-26 sample papers show that dielectric insertion problems (at constant charge vs constant voltage) appear every year for 3 marks.
- Energy stored formula — all three forms: Know \( U = \frac{1}{2}CV^2 \), \( U = \frac{Q^2}{2C} \), and \( U = \frac{1}{2}QV \). Board exams use different given quantities to test flexibility.
- Derivation of parallel plate capacitor: This is a standard 3-mark derivation. Our experts suggest writing it at least five times before the board exam.
- Use dimensional analysis: Always verify your answer’s unit. Capacitance must come out in Farads. A wrong unit signals an algebraic error.
Common Mistakes to Avoid
- Confusing series and parallel rules: Many students apply the resistance rule to capacitors. Remember, for capacitors in series, the reciprocals add. For parallel, the capacitances add directly.
- Forgetting to convert units: Plate separation is often given in mm or cm. Always convert to metres before substituting into \( C = \varepsilon_0 A/d \). Failing to do so gives an answer that is off by a factor of 1000.
- Ignoring the dielectric constant: When a dielectric is inserted, the formula changes from \( C = \varepsilon_0 A/d \) to \( C = K\varepsilon_0 A/d \). Students often forget to multiply by K.
- Mixing up constant charge vs constant voltage: When a battery remains connected, voltage stays constant. When the battery is disconnected, charge stays constant. The energy changes differently in each case. This is a classic JEE trap.
- Using wrong energy formula: Students sometimes write \( U = CV^2 \) instead of \( U = \frac{1}{2}CV^2 \). The factor of one-half is essential and is a common 1-mark error in board exams.
JEE/NEET Application of Capacitance Formula
In our experience, JEE aspirants encounter the Capacitance Formula in at least 2–3 questions per paper. These questions test conceptual depth, not just formula recall. Here are the three most common application patterns:
Pattern 1: Dielectric Insertion (Battery Connected vs Disconnected)
This is the most frequently tested JEE concept. When a dielectric of constant K is inserted:
- Battery connected (V constant): Charge increases to KQ, energy increases to KU.
- Battery disconnected (Q constant): Voltage decreases to V/K, energy decreases to U/K.
JEE 2023 and 2024 both featured variations of this concept. Mastering it can secure 4 marks per question.
Pattern 2: Capacitor Network Simplification
Complex networks of capacitors are reduced using series and parallel rules. NEET frequently tests 3–4 capacitor networks. The key skill is identifying which capacitors share the same two nodes (parallel) and which carry the same charge (series).
\[ C_{series} = \frac{C_1 C_2}{C_1 + C_2} \quad \text{(for two capacitors)} \]
Pattern 3: Energy Methods in Capacitor Problems
JEE Advanced problems often ask for the heat dissipated when capacitors are connected. The key relation is:
\[ \text{Heat dissipated} = U_i – U_f = \frac{1}{2}C_1V_1^2 + \frac{1}{2}C_2V_2^2 – \frac{1}{2}(C_1+C_2)V_f^2 \]
where \( V_f = (C_1V_1 + C_2V_2)/(C_1+C_2) \) is the common final voltage. In our experience, this pattern appears in JEE Advanced roughly once every two years. Practising it thoroughly can be a significant score booster.
FAQs on Capacitance Formula
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